Alternative proof that the sum of two convergent series is convergent

Well, we only want to show that $$ \sum_{n = 1}^\infty \lvert a_n + b_n \rvert < \infty. $$ By the triangle inequality we have: $$ \sum_{n = 1}^\infty \lvert a_n + b_n \rvert \leq \sum_{n = 1}^\infty (\lvert a_n \rvert + \lvert b_n \rvert) $$ Since both $$ \sum_{n = 1}^\infty \lvert a_n \rvert \quad \text{ and } \quad \sum_{n = 1}^\infty \lvert b_n \rvert $$ converge, we can rewrite $$ \sum_{n = 1}^\infty (\lvert a_n \rvert + \lvert b_n \rvert) = \sum_{n = 1}^\infty \lvert a_n \rvert + \sum_{n = 1}^\infty \lvert b_n \rvert < \infty. $$ So $$ \sum_{n = 1}^\infty \lvert a_n + b_n \rvert < \infty $$ according to the direct comparison test.

In general: If we can prove that some series $\displaystyle \sum_{n = 1}^\infty \lvert c_n \rvert$ is bounded, we know (at least in $\mathbb{R}$) that it converges, since the partial sums $$ S_m: m \mapsto \sum_{n = 1}^m \lvert c_n \rvert $$ are bounded and increasing. To see the latter just note that ($m \in \mathbb{N})$: $$ 0 \leq \lvert a_{m+1} - b_{m+1} \rvert \implies S_m \leq S_{m} + \lvert a_{m+1} - b_{m+1} \rvert = S_{m+1} $$